hdu 4496 并查集 逆向 并查集删边

貌似某大犇说过 正难则反，，，

题目说要对这张图进行删边，然后判断联通块的个数，那么就可以先把所有边都删掉，之后从后往前加边，若加的边两端点不在同一个联通块中，

那么此时联通快个数少一，否则不变

#include <cstdio>
#include <cstring>
#include <algorithm>

const int maxn = 10000 + 500;
const int maxm = 100000 + 5000;
int father[maxn];
int x1[maxm], x2[maxm];
int ans[maxm];
int n, m;


int getfather(int x) {
    if (father[x] == x) return (x);
    return (father[x] = getfather(father[x]));
}

int main () {
    while(scanf("%d %d", &n, &m) != EOF) {
        memset(ans, 0, sizeof(ans));
        for (int i = 1; i <= n; i++) father[i] = i;
        for (int i = 1; i <= m; i++) {
            scanf("%d %d", &x1[i], &x2[i]);
            x1[i] += 1;
            x2[i] += 1;
        }
        ans[m] = n;
        for (int i = m; i >= 1; i--) {
            int tx = getfather(x1[i]);
            int ty = getfather(x2[i]);
            if (tx != ty) {
                ans[i-1] = ans[i] - 1;
                father[tx] = ty;
            } else {
                ans[i-1] = ans[i];
            }
        }
        for (int i = 1; i <= m; i++) printf("%d
", ans[i]);
    }
    return 0;
}